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Topic: Contour integral
Replies: 3   Last Post: May 26, 2012 11:23 AM

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Jim Heckman

Posts: 139
Registered: 9/29/06
Re: Contour integral
Posted: May 26, 2012 12:53 AM
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On 25-May-2012, Robin Chapman <R.J.Chapman@ex.ac.uk>
wrote in message <jpnjar$lrp$1@dont-email.me>:

> > I(a,b) = Int_{-inf}^{+inf} exp[-a^2*(x+b)^2] dx
> >
> > where a and b are complex and Re(a^2)> 0 (so the integral
> > converges).
> >
> > The text claims without proof that:
> >
> > The method of residues enables us to show that this integral
> > does not depend on b:
> >
> > I(a,b) = I(a,0)

>
> Integrate exp(-a^2 z^2) over rectangular contour: vertices +-R,
> +-R + i Im(b).
>

> > and that, when the condition -pi/4< Arg(a)< +pi/4 is
> > fulfilled (which is always possible if Re(a^2)> 0), I(a,0)
> > is given by:
> >
> > I(a,0) = I(1,0)/a = sqrt(pi)/a

>
> Integrate exp(-z^2) over "slice-of-pie" contour: centre 0
> vertices at aR and |a|R.


Thanks, Robin. And may I say it's nice to see you here again in
<sci.math>. I've noticed you've poked your nose in once or twice
recently -- hope you stick around awhile.

--
Jim Heckman



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